where r is the position vector from the current length element to the point P.

(i)

A circular loop of wire of radius R lies in the xy-plane and is centred at the origin.
The wire carries a steady current I, which is seen to be flowing clock-wise round the
loop when looking along the z-direction from the negative z towards the positive z.
Calculate the magnetic field at a distance Z on the axis of the circular loop and show
that the z-component of the field is given by:

[tex] B_z = \frac{\mu_oIR^2}{2(Z^2+R^2)^{3/2}} [/tex]

(ii)

A single-turn, circular loop of radius 10 cm is to produce a field at its centre that
will just cancel the earth’s magnetic field at the equator, which is 0.7 G directed
north. Find the current in the loop (1 T ≡ 104 Gauss).

(iii)

Make a sketch, showing the orientation of the loop and the current for part (ii).

[tex] \vec{r} [/tex] is a vector, but I am not sure of what it represents...?

???

TFM

You said it yourself in your first post:

[tex] \vec{r} [/tex] is the position vector from the current length element to the point P.

draw a picture, and try to express this in terms of cylyndrical coordinates and unit vectors....remember, since you are trying to find the field at "a distance Z on the axis of the circular loop", your point P is at [itex](r, \phi, z)=(0,0,Z)[/itex]

draw a picture, and try to express this in terms of cylyndrical coordinates and unit vectors....remember, since you are trying to find the field at "a distance Z on the axis of the circular loop", your point P is at [tex] (r, \phi, z)=(0,0,Z) [/tex]

???

TFM

Attached Files:

The second one is close, but you have mislabeled 'r'...the radius of the circular path of the wire is capital 'R', little 'r' should represent the distance form P to an infinitesimal length of the wire dl....you should shade in a short section of the wire and call it 'dl'...then find an expression for the vector from 'dl' to 'P'...

Okay so firstlty I have the midifed attached image (Its a JPEG, hence the reason the filling in is poor)

so I need an expression for the vector from dl to P. would it not be [tex] \hat{z} [/tex], since the point P is on the Z axis, and so the vector would go from the centre of the loop to P, which is just up along the Z-Axis

Okay, call that angle [itex]\alpha[/itex]; then the direction of [itex]\vec{r}[/itex] is [itex]-\cos \alpha \hat{r}+\sin \alpha \hat{z}[/itex] Is it not?...now look at the triangle, what are [itex]\cos \alpha[/itex] and [itex]\sin \alpha[/itex] in terms of R and Z?